Question

Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of its chemical process from the historical process mean of 750 pounds per hour. When five trial runs are made using the new catalyst, the following yields (in pounds per hour) are recorded: 801, 814, 784, 836, and 820. ChemYield

t-statistic
6.942585
p-value
0.002261

Answer 1

Answer:

Null hypothesis:$\mu = 750$  

Alternative hypothesis:$\mu \neq 750$  

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$  

$p_v =2*P(t_{4}>6.943)=0.00226$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

Step-by-step explanation:

Data given and notation

Data:    801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=811$ represent the sample mean  

$s=19.647$ represent the standard deviation for the sample

$n=5$ sample size  

$\mu_o =750$ represent the value that we want to test  

$\alpha$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:  

Null hypothesis:$\mu = 750$  

Alternative hypothesis:$\mu \neq 750$  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$  

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

What do you conclude?  

Compute the p-value  

Since is a two tailed test the p value would be:  

$p_v =2*P(t_{4}>6.943)=0.00226$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

Answer 2

Considering the p-value of the test, it is found that since 0.002261 < 0.05, there is enough evidence to conclude that the new catalyst changes the mean hourly yield of its chemical process from the historical process mean of 750 pounds per hour.

What are the hypothesis tested?

At the null hypothesis, it is tested if the mean is not changed, that is:

$H_0: \mu = 750$.

At the alternative hypothesis, it is tested if the mean is changed, that is:

$H_1: \mu \neq 750$.

What is the decision?

The decision depends on the p-value, as follows:

• If p-value < significance level, the null hypothesis is rejected.

• If p-value > significance level, the null hypothesis is not rejected.

In this problem, we consider a standard significance level of 0.05, hence, since 0.002261 < 0.05, we reject the null hypothesis, that is, there is enough evidence to conclude that the new catalyst changes the mean hourly yield of its chemical process from the historical process mean of 750 pounds per hour.

More can be learned about p-values at https://brainly.com/question/16313918